It is no doubt desirable to illustrate Eudoxus’s method by one example. We will take one of the simplest, the proposition (Eucl. XII., 10) about the cone. Given ABCD, the circular base of the cylinder which has the same base as the cone and equal height, we inscribe the square ABCD; we then bisect the arcs subtended by the sides, and draw the regular inscribed polygon of eight sides, then similarly we draw the regular inscribed polygon of sixteen sides, and so on. We erect on each regular polygon the prism which has the polygon for base, thereby obtaining successive prisms inscribed in the cylinder, and of the same height with it. Each time we double the number of sides in the base of the prism we take away more than half of the volume by which the cylinder exceeds the prism (since we take away more than half of the excess of the area of the circular base over that of the inscribed polygon, as in Euclid XII., 2). Suppose now that V is the volume of the cone, C that of the cylinder. We have to prove that C = 3V. If C is not equal to 3V, it is either greater or less than 3V.
Suppose (1) that C > 3V, and that C = 3V + E. Continue the construction of prisms inscribed in the cylinder until the parts of the cylinder left over outside the final prism (of volume P) are together less than E.
| Then | C − P < E. |
| But | C − 3V = E; |
| Therefore | P > 3V. |
But it has been proved in earlier propositions that P is equal to three times the pyramid with the same base as the prism and equal height.
Therefore that pyramid is greater than V, the volume of the cone: which is impossible, since the cone encloses the pyramid.
Therefore C is not greater than 3V.
Next (2) suppose that C < 3V, so that, inversely,
V > 1⁄3 C.
This time we inscribe successive pyramids in the cone until we arrive at a pyramid such that the portions of the cone left over outside it are together less than the excess of V over 1⁄3 C. It follows that the pyramid is greater than 1⁄3 C. Hence the prism on the same base as the pyramid and inscribed in the cylinder (which prism is three times the pyramid) is greater than C: which is impossible, since the prism is enclosed by the cylinder, and is therefore less than it.